i i 0 1 534b this formula makes sense when

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Unformatted text preview: 3.4c) Substituting (5.3.4c) into (5.3.3a,b) yields Pk;1(t) = k ;1 X (;1)i ; i i=0 14 rifn;1 (5.3.5a) and, using (5.3.1), Ek;1(t) = (;1)k hk ; k y(k+1)( ): (5.3.5b) Replacing f (t y(t)) in (5.3.2) by Pk;1(t) + Ek;1(t) and transforming the integrals from t to using (5.3.4a) yields y(tn) = y(tn;1) + h Z 1X k ;1 0 i=0 ; (;1)i rifn;1 d i Let m = (;1)m Z ; 1 y(tn) = y(tn;1) + h i=0 i ri f n;1 + (;1)k hk+1 1 0 (;1)k hk ; k d: m 0 Then we have k;1 X +h Z Z 1 (5.3.6a) ; 0 y(k+1)( )d : k y(k+1)( )d : (5.3.6b) Dropping the error term yields the k-step Adams-Bashforth method yn = yn;1 + h The local error of this formula is dn = (;1)k hk+1 Z k;1 X i=0 1 ; k 0 i i r fn;1 : y(k+1)( )d : (5.3.7a) (5.3.7b) Comparing (5.3.7a) with the general multistep formula (5.1.2a), we see that 0 = 1, 1 = ;1, and 2 = 3 = : : : = k = 0 = 0. Thus, (5.3.7a) uses the one solution value yn;1 and the k function values fn;1, fn;2, : : : , fn;k to obtain the yn (cf. Figure 5.3.1). We'll call (5.3.7a) a k-step method since it uses solution and function values at the k points tn;1 , tn;2 , : : : , tn;k however, other terminology is also used 10]. From (5.3.7b) we see that the local error of (5.3.7a) is O(hk+1). Thus, the order of a k-step Adams-Bashforth method is k. The method (5.3.7) has only one function evaluation per step. This is contrasted to the minimum of k function evaluations that are needed for a k-stage Runge-Kutta method. 15 y y(t) f n-2 f n-k t n-k t n-2 yn-1 f n-1 t n-1 tn t Figure 5.3.1: Information needed for a k-step Adams-Bashforth method. The representation of (5.3.7a) in terms of backward di erences will be useful to us in some circumstances, but generally we prefer a formula written in terms of function values at preceding points. This can be done using q X jq qf r n;1 = (;1) j fn;1;j : (5.3.8a) j =0 Then, using (5.3.7a) yn = yn;1 + h k;1 i XX i=0 i j =0 (;1)j if j n;1;j : (5.3.8b) By rearranging the order of the computation (5.3.8b) can be written in the form (Problem 1) yn = yn;1 + h where j = k;1 X i=j ;1 i k...
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